Medium Security
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For solving https://sudokupad.app/1yafhqebza

1. R7C7 is larger than 6, but isn't 7 or 9, so it's 8. That, along with given 8s, forces 8 onto row 8 in box 8, and into column 9 in box 6. The given 8s in boxes 3 and 7 force 8 into one of three cells in box 1, but either of the white cells would require a digit larger than 8 in the red cell, and 9 is already in the box, so R3C3 is 8. That forces 8 into row 2 in box 2, and column 2 in box 4.

2. Since box 2 has an 8 in column 2, that requires a larger digit in the adjacent red cell, so there is a 9 in column 3 in box 2. Combined with the given 9 in box 1, that means R7C1 is 9. Combined with the given 9 in box 9, that forces 9 into row 9 in box 8. In box 6, 9 cannot be in column 9, and it also cannot be in column 8, because that would require a digit larger than 9 in the adjacent red call, so 9 is in column 7 in box 6. That forces R3C8 to be 9, and 9 in box 2 is in row 2.

3. From high to low: 1 is in column 2 in box 1, and there are several ways to see where 1 goes in box 4. It can't be in column 3, because those are red cells, and there's no digit smaller than 1 to go next to them. It also can't be in column 3 because that would leave no place for 1 in column 1. So 1 in box 4 is in column 1, and in box 7, it's in column 3. But not in R8C3, because there's no digit smaller than 1 to go into the pink cell. In box 8, there are too many options for marking, but we can see 1 can't go into row 8, because there would need to be a digit smaller than 1 in the adjacent pink cell. That leave only one place for 1 in row 8, so R8C7 is 1.

4. We can finish box 9: the missing digits are 2 and 5, and the given 5 in box 7 means R9C7 is 2 and R7C9 is 5.

5. There are three digits missing in column 8, so they're a 157 triple. Combined with the placed 1 in box 9, that forces R3C9 to be 1.

6. More triples: column 9 needs 248. The three red cells in box 6 are a 369 triple. The three cells in box 3 are a 457 triple, but there's no 5 in R1C7, and no 7 in R2C7. Pink cells must be smaller than non-pink neighboring cells, so R3C7 is 4, R2C7 is 5, and R1C7 is 7.

7. Now 5 and 7 are in pink cells in box 2, while 1 is in row 1. In box 1, 4 can't be in row 3, so it's in column 3. That means there's a 4 in column 1 in box 4, and in row 7 in box 8. Also in box 8, there's a 5 in row 8. There's also a 5 in column 3 in box 4.

8. The given 6 in box 9 means the 6 in box 7 is in column 3. That means 6 in box 1 is in row 3. That means 6 in box 2 is in row 2, making a 689 triple, and forcing R2C3 to be 4.

9. Every cell in box 5 other than the center is adjacent to either a pink or red cell. 9 can't be next to a red cell, and 1 can't be next to a pink cell, forcing 9 into column 5, and 1 into row 5. Let's consider the effect of that on box 6. Specifically on the red cells in box 6: a 369 triple. Each of those red cells is larger than the adjacent cells in columns 6 and 8, and we know that column 8 contains a 157 triple. These can be, and in fact must be, paired up: 7 must be next to 9, 5 next to 7, and 1 next to 3. Since 3 is not larger than 4, R4C7 is not 3 and R4C8 is not 1. Since the 1 in box 5 must be on row 5, R5C8 is also not 1, and R5C7 is also not 3. That means R6C8 is 1 and R6C7 is 3. The only other digit smaller than 3 has to go on the other side, so R6C6 is 2. R7C6 is smaller, and so is 1.

10. Those placed digits affect some of our earlier pencil marks: R9C3 is now 1, and R9C3 is now 6. We can finish box 7 using the pink cell: R7C2 is 7 and R7C3 is 3. We can finish out the pencil marks in box 8 now: the pink cells are a 24 pair. Row 8 is a 578 triple. Row 9 is a 369 triple, although R9C5 is not a 9.

11. Column 3 is missing 2, 5, 7, and 9, but R1C3 already shares a row with 5, 7, and 9, so it is 2, and the red cells in box 4 are a 579 triple. The two missing digits in box 1 are a 36 pair, and the missing digits in row 1 are a 134 triple, although R1C6 is not 1. The pink cells in box 2 are a 257 triple, although R3C6 is not 2. Since R3C6 is either 5 or 7, and must be smaller than R4C6, which in turn must be smaller than R4C7, the minimum value of R4C7 is 7. Since that cell is marked to be either 6 or 9, R4C7 is 9, and R5C7 is 6, so R5C8 is 5, and R4C8 is 7.

12. The just-placed 7 and 9 on row 4 force R4C3 to be 5. R4C2 is either 2 or 3 because of the givens in the column. R4C4 is either 3 or 4 because of the placed 2 in box 5 and the pink cells forcing 1 into row 5. R3C4 must be at least two digits lower than 5, so it is 2. That forces R7C4 to be 4, and R7C5 to be 2. That bounces back, so R4C4 is 3, R4C2 is 2, and R1C4 is 1. This affects more pencil marks, and now R5C5 is 1, and R6C5 is 9. That affects column 3: R6C3 is 7, and R5C3 is 9. More pencil marks resolved: R4C1 is 1, and R5C9 is 2.

13. 7 in box 5 has to be on row 5, and smaller than the adjacent red cell, so R5C4 is 7. The three missing digits in row 5 are a 348 triple, but R5C6 can't be 3 (because of the 3 in the box), nor 8 (becauuse it must be smaller than 6), so R5C6 is 4. That forces R1C6 to be 3 and R1C5 to be 4. To finish out row 5 we need 3 and 8, and the given 8 means R5C1 is 3 and R5C2 is 8. That resolves pencil marks in box 1, so R3C1 is 6 and R3C2 is 3. Then R6C1 is 4 and R6C2 is 6. After that R6C9 is 8 and R6C4 is 5.

14. The two empty cells in box 5 are a 68 pair, so R4C9 is 4. Column 5 now also has a 68 pair, so the digit that's lower than both of them in R3C5 is 5, and R3C6 is 7. The digits that are both larger than 7 in column 6 are R2C6, which is 9, and R4C6, which is 8.

15. Cells are collapsing now: R4C5 is 6, R2C5 is 8, and R2C4 is 6.

16. The remaining six cells have only one possible value each.